Why Read This Book? - Index of

174 Chapter 5 The Real Numbers
EXERCISE 5.3.11 Let F ={A} be a family of closed sets, and let {Bk} n k=1 be a
finite family of closed sets. Then
(a) �
F A is closed.14
(b) � n k=1 Bk is closed.
By Exercises 5.3.9 and 5.3.11, it follows that any finite set is closed because it
is a finite union of singleton sets. The integers are a good example of an infinite
set that is closed, because Z C =∪n∈Z(n, n + 1).
EXERCISE 5.3.12 Give an example of an infinite family F of closed sets such
that ∪F A is not closed. Prove. 15
EXERCISE 5.3.13 Show that {1/n} ∞ n=1 is not closed by showing its complement
is not open.
Lots of sets are neither open nor closed. For example, [a, b) is sometimes called
a half-open or half-closed interval. Believe it or not, some sets are both open and
closed. The empty set and the real numbers are both open and closed because
they are both open and they are complementary. Theorem 5.3.14 uses the LUB
axiom to show the important fact that these are the only two sets of real numbers
that are both open and closed. The proof is a little intricate, so we provide it here.
It is important for you to work your way through the details because it illustrates
an important type of proof where we show that certain examples of something are
the only ones that exist. Keep a pencil and paper handy to sketch some number
lines and help you visualize the details.
Theorem 5.3.14 The empty set and the real numbers are the only subsets of
the real numbers that are both open and closed.
Proof. We prove by contradiction. Suppose there exists a non-empty proper subset
A of the real numbers that is both open and closed. Then A and A C are both
open and non-empty. Pick any a1 ∈ A and a2 ∈ A C . Without any loss of generality,
we may assume a1 0 : Nɛ(a1) ⊆ A}; that is, E is the set of all
ɛ>0 such that the ɛ-neighborhood of a1 is contained entirely within A. Since A
is open, E is non-empty. Because a2 ∈ A C , E is bounded from above by a2 − a1.
Thus we may apply the LUB property to E to conclude the existence of some ɛ0,
the LUB of E. Thus Nɛ0 (a1) is the largest neighborhood of a1 that is contained
entirely within A.
In order to arrive at a contradiction, we look at the end points of the interval
(a1 − ɛ0,a1 + ɛ0). First we ask if it is possible that a1 + ɛ0 ∈ A C . If so, then the fact
14 See Exercise 3.3.8.
15 You shouldn’t have to look too far.